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24 tháng 3 2020
a) Ta có: \(\frac{x^3-3x^2+x-3}{x-3}\)
\(=\frac{x^2\left(x-3\right)+\left(x-3\right)}{\left(x-3\right)}=\frac{\left(x-3\right)\left(x^2+1\right)}{x-3}=x^2+1\)
b) Ta có: \(\frac{x^2+2x+x^2-4}{x+2}\)
\(=\frac{x\left(x+2\right)+\left(x+2\right)\left(x-2\right)}{x+2}=\frac{\left(x+2\right)\left(x+x-2\right)}{x+2}=2x-2\)
c) Ta có: \(\frac{2x^3-5x^2+6x-15}{2x-5}\)
\(=\frac{x^2\left(2x-5\right)+3\left(2x-5\right)}{2x-5}=\frac{\left(2x-5\right)\left(x^2+3\right)}{2x-5}=x^2+3\)
TT
1 tháng 5 2018
a \(2x+2>4\\ \Leftrightarrow2\left(x+1\right)>4\\ \Leftrightarrow x+1>2\\ \Leftrightarrow x>1\)
b \(3x+2>-5\\ \Leftrightarrow3x>-7\\ \Leftrightarrow x>\dfrac{-7}{3}\)
c \(10-2x>2\\ \Leftrightarrow2\left(5-x\right)>2\\ \Leftrightarrow5-x>1\\ \Leftrightarrow-x>-4\\ \Leftrightarrow x< 4\)
d \(1-2x< 3\\ \Leftrightarrow-2x< 2\\ \Leftrightarrow2x>2\\ \Leftrightarrow x>1\)
KT
4 tháng 5 2018
a)2x+2>4
<=> 2x>4-2
<=>2x>2
<=>x>1
Vậy...
b)3x+2>-5
<=>3x>-5-2
<=>3x>-7
<=>x>\(\dfrac{-7}{3}\)
Vậy...
c)10-2x>2
<=>-2x>-10+2
<=>-2x>-8
<=>x<4
Vậy...
d)1-2x<3
<=>-2x<3-1
<=>-2x<2
<=>x>-1
Vậy...
e)10x+3-5\(\le\)14x+12
<=>10x-2\(\le\)14x+12
<=>10x-14x\(\le\)2+12
<=>-4x\(\le\)14
<=>x\(\ge\)\(\dfrac{-7}{2}\)
Vậy...
f)(3x-1)<2x+4
<=> 3x-2x<1+4
<=>x<5
Vậy...
24 tháng 5 2022
a: \(\left(x-1\right)\left(x-2\right)\left(x-3\right)\)
\(=\left(x^2-3x+2\right)\left(x-3\right)\)
\(=x^3-3x^2-3x^2+9x+2x-6\)
\(=x^3-6x^2+11x-6\)
b: \(\left(x^2+x+1\right)\left(x^2-1\right)\left(x^2-x+1\right)\)
\(=\left(x-1\right)\left(x^2+x+1\right)\cdot\left(x+1\right)\left(x^2-x+1\right)\)
\(=\left(x^3-1\right)\left(x^3+1\right)\)
\(=x^6-1\)
c: \(=8x-6x^2-20+15x-\left(15x-6x^2+55-10x\right)-30x+75\)
\(=-6x^2-7x+55+6x^2-5x-55\)
\(=-12x\)
d: \(\left(x^2-2x+3\right)\left(3x-5\right)-\left(x^2+x-1\right)\left(2x+7\right)\)
\(=3x^3-5x^2-6x^2+10x+9x-10-\left(x^2+x-1\right)\left(2x+7\right)\)
\(=3x^3-11x^2+19x-10-\left(2x^3+7x^2+2x^2+7x-2x-7\right)\)
\(=3x^3-11x^2+19x-10-2x^3-9x^2-5x+7\)
\(=x^3-20x^2+14x-3\)
BL
14 tháng 3 2020
a. \(1+\frac{2x-5}{6}=\frac{3-x}{4}\)
\(\Leftrightarrow\frac{12}{12}+\frac{2\left(2x-5\right)}{12}-\frac{3\left(3-x\right)}{12}=0\)
\(\Leftrightarrow12+4x-10-9+3x=0\)
\(\Leftrightarrow7x-7=0\)
\(\Leftrightarrow7x=7\Leftrightarrow x=1\)
b. \(\frac{x+1}{2}-\frac{x-2}{3}=\frac{3\left(x+1\right)}{6}-\frac{2\left(x-2\right)}{6}=\frac{x+7}{6}\)
c. \(\frac{2x-1}{3}+x=\frac{x+4}{2}\)
\(\Leftrightarrow\frac{2\left(2x-1\right)}{6}+\frac{6x}{6}-\frac{3\left(x+4\right)}{6}=0\)
\(\Leftrightarrow7x-14=0\)
\(\Leftrightarrow7x=14\Leftrightarrow x=2\)
d. \(\frac{x+5}{4}-\frac{2x-3}{3}-\frac{6x-1}{8}+\frac{2x-1}{12}\)
\(=\frac{6\left(x+5\right)}{24}-\frac{8\left(2x-3\right)}{24}-\frac{3\left(6x-1\right)}{24}+\frac{2\left(2x-1\right)}{24}\)
\(=6x+30-16x+24-18x+3+4x-2\)
\(=-24x-55\)
=>2x-4|x-2|=3x+15
=>4|x-2|=-x-15
TH1: x>=2
=>4x-8=-x-15
=>5x=-7
=>x=-7/5(loại)
TH2: x<2
=>4x-8=x+15
=>3x=23
=>x=23/3(loại)